Midpoint of a line segment

Using the midpoint formula , the midpoint of segment $PQ$ will be $\left(\dfrac{x+1}{2},\dfrac{y+1}{2}\right)$ . The product is $36$ , so we have

$\left(\dfrac{x+1}{2}\right) \left(\dfrac{y+1}{2}\right)=36$ $\implies$ $\dfrac{(x+1)(y+1)}{4}=36$ $\implies$ $(x+1)(y+1)=144$

To maximize $x$ , we let the term $x+1$ equal to the largest factor of $144$ , so $x+1=144$ and $y+1=1$ . However, this implies that $y=0$ , which is not a positive integer. Therefore, we let $x+1$ equal to the next largest factor of $144$ which is $72$ . Then we have

$x+1=72$ and $y+1=2$

It follows that $x=71$ and $y=1$ .

Therefore, the largest possible value of $x$ is $\boxed{71}$ .